Differential equations and Laplace transforms: Laplace-transformations
Transfer and response functions
We summarize the use of the Laplace transformation by describing a general method for finding specific solutions of second-order linear equations with constant coefficients.
Consider the initial value problem \[a\cdot x''+b\cdot x'+c\cdot y = g(t)\qquad\text{ with }\qquad x(0) = x_0\quad \text{ and }\quad x'(0)=x'_0\]Let
- \(H(s) = \frac{1}{a\cdot s^2+b\cdot s+c}\) be the transfer function, and
- \(h=\mathcal{L}^{-1}(H)\) the impulse response function of the IVP.
In terms of these functions, the solution of the above IVP is the sum of the following two functions:
- the forced response: the convolution \(g\ast h\); it is the solution of the IVP \(a\cdot x''+b\cdot x'+c\cdot y = g(t)\) with \(x(0)=x'(0)=0\)
- the free response: \( \mathcal{L}^{-1}\left(H(s)\cdot\left((a\cdot s+b)\cdot x_0+a\cdot x_0'\right)\right)\); it is the solution of the IVP \(a\cdot x''+b\cdot x'+c\cdot y = 0\) with \(x(0)=x_0\) and \(x'(0)=x_0'\)
The example below shows how the above transfer and response functions help in solving a second-order linear IVP.
\[ {{d^2}\over{d t^2}} x-10 {{d}\over{d t}} x+9 x=5 t\qquad \text{ with }\qquad x(0)=-1\ \text{ and }\ x'(0)=2\]
\(x(t)=\) \( {{31\cdot \euler^{9\cdot t}-162\cdot \euler^{t}+45\cdot t+50}\over{81}} \)
We begin by determining the transfer function #H(s)# of the IVP. By definition of the transfer function, we have
\[\begin{array}{rcl}
H(s) &=&\dfrac{1}{as^2+bs+c}\\
&&\phantom{xx}\color{blue}{\text{if }a,\ b,\ c\ \text{are constants such that the ODE is of the form}}\\
&&\phantom{xx1234567891011}\color{blue}{ax''+bx'+cx = g(t)}\\
&=&\displaystyle {{1}\over{s^2-10 s+9}}
\end{array}\]
Thus, \(H(s) =\) \( {{1}\over{s^2-10 s+9}} \).
We next determine the impulse response function \(h(t)\) of this problem. By definition of the impulse response function and the Inverse Laplace transforms of rational functions, we have
\[\begin{array}{rcl}
h(t) &=&\mathcal{L}^{-1}(H(s))(t)\\
&=&\displaystyle \mathcal{L}^{-1}\left({{1}\over{s^2-10 s+9}}\right)(t)\\
&&\phantom{xx}\color{blue}{\text{expression found for }H(s)\text{ filled in}}\\
&=&\displaystyle {{\euler^{9\cdot t}}\over{8}}-{{\euler^{t}}\over{8}}\\
&&\phantom{xx}\color{blue}{\text{the inverse Laplace transform of a rational function}}\\
\end{array}\]
Thus, \(h(t) =\) \( {{\euler^{9\cdot t}}\over{8}}-{{\euler^{t}}\over{8}} \).
The Laplace transform of the free reponse is
\[H(s)\cdot\left((a\cdot s+b)\cdot x_0+a\cdot x_0'\right) = {{12-s}\over{s^2-10\cdot s+9}}\]
The free reponse itself is the inverse Laplace transform of this function:
\[ \mathcal{L}^{-1}\left(H(s)\cdot\left((a\cdot s+b)\cdot x_0+a\cdot x_0'\right)\right) = {{3\cdot \euler^{9\cdot t}}\over{8}}-{{11\cdot \euler^{t}}\over{8}}\]
In order to determine the forced response, we compute the following convolution product:
\[\begin{array}{rcl} (g\ast h)(t) &=&\displaystyle \int_0^t h(\tau)\cdot g(t-\tau)\,\dd \tau\\
&=&\displaystyle \int_0^t {{-\left(5\cdot \tau-5\cdot t\right)\cdot \euler^{9\cdot \tau}-\left(5\cdot t-5\cdot \tau\right)\cdot \euler^{\tau}}\over{8}}\,\dd \tau\\
&=&\displaystyle \left[ {{-\left(45\cdot \tau-45\cdot t-5\right)\cdot \euler^{9\cdot \tau}-\left(-405\cdot \tau+405\cdot t+405\right)\cdot \euler^{\tau}}\over{648}}\right]_{\tau=0}^{\tau=t}\\ &=&\displaystyle{{360\cdot t+400}\over{648}}-{{405\cdot \euler^{t}-5\cdot \euler^{9\cdot t}}\over{648}}\\ &=&\displaystyle {{5\cdot \euler^{9\cdot t}-405\cdot \euler^{t}+360\cdot t+400}\over{648}}
\end{array}\]
Adding the free and forced response, we find the solution of the IVP:
\[\begin{array}{rcl} x(t)& =& \displaystyle {{3\cdot \euler^{9\cdot t}}\over{8}}-{{11\cdot \euler^{t}}\over{8}}+{{5\cdot \euler^{9\cdot t}-405\cdot \euler^{t}+360\cdot t+400}\over{648}}\\ &=& \displaystyle {{31\cdot \euler^{9\cdot t}-162\cdot \euler^{t}+45\cdot t+50}\over{81}}\end{array}\]
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